You are given several identical resistors each of value $5 Ω$ and each capable of carrying a maximum current of 2 A. It is required to make a suitable combination of these resistances to produce a resistance of $2 .5 Ω$ which can carry current of 4 A. The minimum number of resistances required for this job is

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Detailed Solution

$R = 5 Ω R_{eff} = 2 .5$
When two resistors are connected parallel, its effective resistance becomes $2 .5 Ω$
$R_{eff} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{5 \times 5}{5 + 5} = 2 .5 Ω$
This combination can carry a current of 4A

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